Problem: Solve for $r$, $ \dfrac{r - 5}{2r - 2} = -\dfrac{5}{4r - 4} + \dfrac{4}{5r - 5} $
Explanation: First we need to find a common denominator for all the expressions. This means finding the least common multiple of $2r - 2$ $4r - 4$ and $5r - 5$ The common denominator is $20r - 20$ To get $20r - 20$ in the denominator of the first term, multiply it by $\frac{10}{10}$ $ \dfrac{r - 5}{2r - 2} \times \dfrac{10}{10} = \dfrac{10r - 50}{20r - 20} $ To get $20r - 20$ in the denominator of the second term, multiply it by $\frac{5}{5}$ $ -\dfrac{5}{4r - 4} \times \dfrac{5}{5} = -\dfrac{25}{20r - 20} $ To get $20r - 20$ in the denominator of the third term, multiply it by $\frac{4}{4}$ $ \dfrac{4}{5r - 5} \times \dfrac{4}{4} = \dfrac{16}{20r - 20} $ This give us: $ \dfrac{10r - 50}{20r - 20} = -\dfrac{25}{20r - 20} + \dfrac{16}{20r - 20} $ If we multiply both sides of the equation by $20r - 20$ , we get: $ 10r - 50 = -25 + 16$ $ 10r - 50 = -9$ $ 10r = 41 $ $ r = \dfrac{41}{10}$